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3 years ago

In trapezoid KLMN, solve for x ,then use that value to solve for

Mathematics

1 answer:

Andre45 [30]3 years ago

3 0

Answer:

<u>Same side angles are supplementary</u>

L+k=180

18x-18+72=180

18x=126

x=7

m∠18(7)-18

= 108°

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hope it helps...

have a grat day

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3 years ago

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3 years ago

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Triss [41]

Answer:

The roots of f(x) are: -2, 3, (1+3i) and (1-3i)

Step-by-step explanation:

We are given an expression:

$f(x)=x^4-3x^3+6x^2+2x-60$

(1+3i) is a root of f(x)

We have to find the remaining roots of f(x).

Since, (1+3i) is a root of f(x),

$x-(1+3i)$

is a factor of given expression.

Now, we check if (1 - 3i) is a root of given function.

$f(x)=x^4-3x^3+6x^2+2x-60\\f(1-3i)=(1-3i)^4-3(1-3i)^3+6(1-3i)^2+2(1-3i)-60\\= (28+96i)-3(-26+18i)+6(-8-6i)+2(1-3i)-60 = 0$

Thus, (1-3i) is also a root of given function.

Since, (1-3i) is a root of f(x),

$x-(1-3i)$

is a factor of given expression.

Thus, we can write:

$(x-(1+3i))(x-(1-3i))\text{ is a factor of f(x)}\\x^2-x(1-3i)+x(1+3i)+(1-3i)(1+3i)\text{ is a factor of f(x)}\\x^2-2x+10\text{ is a factor of f(x)}$

Dividing f(x) by above expression:

$\displaystyle\frac{x^4-3x^3+6x^2+2x-60}{x^2-2x+10} = x^2-x-6$

To find the root, we equate it to zero:

$x^2-x-6 = 0\\x^2 - 3x+2x-6=0\\x(x-3)+2(x-3)=0\\(x+2)(x-3) = 0\\x = -2, x = 3$

Thus, the roots of f(x) are: -2, 3, (1+3i) and (1-3i)

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4 years ago